hdu 3549 Flow Problem 【最大流】

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<iostream>
 4 #include<algorithm>
 5 #include<queue>
 6 using namespace std;
 7 
 8 const int maxn = 10005;
 9 const int INF = (1 << 30) - 1;
10 
11 struct Edge{
12     int v,next,c;
13 }e[maxn];
14 
15 int st,ed,lev[maxn],first[maxn],now[maxn],ecnt;
16 int n,m;
17 
18 void init(int a,int b){
19     st = a; ed = b;
20     memset(first,-1,sizeof(first));
21     ecnt = 0;
22 }
23 
24 void addedges(int u,int v,int c){
25     e[ecnt].next = first[u];
26     e[ecnt].v = v;
27     e[ecnt].c = c;
28     first[u] = ecnt++;
29     
30     e[ecnt].next = first[v];
31     e[ecnt].v = u;
32     e[ecnt].c = 0;
33     first[v] = ecnt++;
34 }
35 
36 bool bfs(){
37     queue<int> q;
38     while(!q.empty()) q.pop();
39     q.push(st);
40     memset(lev,-1,sizeof(lev));
41     lev[st] = 0;
42     while(!q.empty()){
43         int x = q.front();q.pop();
44         for(int i = first[x];~i;i = e[i].next){
45             int v = e[i].v;
46             if(lev[v] < 0 && e[i].c > 0){
47                 lev[v] = lev[x] + 1;
48                 q.push(v);
49             }
50         }
51     }
52     return lev[ed] != -1;
53 }
54 
55 int dfs(int p,int minf){
56     if(p == ed || minf == 0) return minf;
57     for(int &i = now[p];~i;i = e[i].next){
58         int v = e[i].v;
59         if(lev[v] == lev[p] + 1 && e[i].c > 0){
60             int d = dfs(v,min(e[i].c,minf));
61             if(d > 0){
62                 e[i].c -= d;
63                 e[i^1].c += d;
64                 return d;
65             }
66         }
67     }
68     return 0;
69 }
70 
71 int dinic(){
72     int max_flow = 0,p1;
73     while(bfs()){
74         memcpy(now,first,sizeof(first));
75         while((p1 = dfs(st,INF)) > 0)
76         max_flow += p1;
77     }
78     return max_flow;
79 }
80 
81 int main(){
82     int T;
83     int kase = 0;
84     scanf("%d",&T);
85     while(T--){        
86         scanf("%d %d",&n,&m);
87         init(1,n);
88         for(int i = 0;i < m;i++){
89             int a,b,c;
90             scanf("%d %d %d",&a,&b,&c);
91             addedges(a,b,c);
92         }
93         printf("Case %d: %d\n",++kase,dinic());
94     }
95     return 0;
96 }